I know there's already a question with a title very similar to this, unfortunately as I understand the OP skips over the part of the proof that is not clear to me.

Let $I$ be a set and $f_\alpha$, $\alpha \in I$ be a collection of lower semicontinuous functions. Show that $g=\sup\limits_{\alpha\in I}\,f_\alpha$ is lower semicontinuous.

My attempt: let $S_f(t)=\{x\in\mathbb{R}^n: f(x)>t\}$ for $t\in\mathbb{R}$. It is easy to see that $f$ is lower semicontinuous if and only if $S_f(t)$ is an open set, now define $$S(t)=\bigcup_\limits{\alpha\in I}S_{f_\alpha}(t).$$ Obviously $S(t)$ is open. I should now show that $S_{g}(t)=S(t)$, which ends the proof.

Now, obviously $S(t)\subset S_g(t)$ because if $f_\alpha(x)>t$ for some $\alpha\in I$ then $g(x)>t$ because $g(x)>f_\alpha(x)$ for all $\alpha\in I$, but I fail to see why $S_g(t)\subset S(t)$, I think I should use the definition of supremum but I don't see how. I'm thinking I should prove it for a particular $t\in \mathbb{R}$ because that seems very untrue for any $t$. Thank you.


Fix any $x \in \Bbb{R}^n$. By definition of $\sup$ you have $$\sup_{\alpha \in I} f_{\alpha} (x) > t \ \ \Longleftrightarrow \ \ \exists\alpha \in I : f_{\alpha} (x) > t$$

Hence $$x \in S_g(t) \ \ \Longleftrightarrow \ \ g(x) > t \ \ \Longleftrightarrow \ \ \exists\alpha \in I : x \in S_{f_{\alpha}} (t) \ \ \Longleftrightarrow \ \ x \in \bigcup_{\alpha \in I} S_{f_{\alpha}} (t)$$

  • $\begingroup$ Thank you for your answer, I don't see why your first affirmation is true $\endgroup$
    – user438666
    Jun 13 '17 at 10:06
  • 1
    $\begingroup$ LHS means that $t$ is not an upper bound of $\{ f_{\alpha} (x) : \alpha \in I\}$ (because $t$ is smaller than the smallest upper bound!). RHS means that $t$ is not an upper bound of $\{ f_{\alpha} (x) : \alpha \in I\}$ (bacause there is an element in the set which is larger than $t$!). So, they are both equivalent to the statement "$t$ is not an upper bound of $\{ f_{\alpha} (x) : \alpha \in I\}$". $\endgroup$
    – Crostul
    Jun 13 '17 at 10:10
  • $\begingroup$ Hi! I'm dealing with a similar problem, but I don't see why $S_f(t)$ is open $\Leftrightarrow$ $f$ is lower continous. May someone explain this to me? Thanks! $\endgroup$
    – pcalc
    May 22 '19 at 7:03
  • $\begingroup$ @pcalc This is a standard fact, see this article for a proof. In any case, you can look up "lower semicontinuous" on Google to learn about it. $\endgroup$
    – Crostul
    May 22 '19 at 18:25

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